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Chapter 9: Problem 86

Butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\), used as the fuel in disposable lighters, reacts with oxygen \(\left(\mathrm{O}_{2}\right)\) to produce \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Suppose \(10.00 \mathrm{~g}\) butane is combined with \(10.00 \mathrm{~g} \mathrm{O}_{2}\). (a) Write a balanced chemical equation for the combustion of butane. (b) Which reactant is limiting? (c) What is the theoretical yield of each product for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) How many additional grams of the limiting reactant are required to run this reaction in a balanced fashion?

Short Answer

Expert verified
The balanced chemical equation for the combustion of butane is: \(2 C_{4}H_{10}(g) + 13 O_{2}(g) \rightarrow 8 CO_{2}(g) + 10 H_{2}O(g)\). The limiting reactant is oxygen (O2). The theoretical yield for each product is 8.466 g CO2 and 4.329 g H2O. There are 7.206 g of excess butane (C4H10) remaining after the reaction. An additional 4.288 g of O2 is required to run the reaction in a balanced fashion.

Step by step solution

01

Write a balanced chemical equation for the combustion of butane.

To write a balanced chemical equation for the combustion of butane, first write the unbalanced chemical equation: C4H10 (g) + O2 (g) -> CO2 (g) + H2O (g) Now balance the equation by adjusting the number of molecules for each compound: 2C4H10 (g) + 13O2 (g) -> 8CO2 (g) + 10H2O (g) The balanced chemical equation is: 2 C4H10 (g) + 13 O2 (g) -> 8 CO2 (g) + 10 H2O (g)
02

Determine the limiting reactant.

To determine the limiting reactant, we must use the initial mass of each reactant: Initially provided: 10.00 g C4H10 and 10.00 g O2 Calculate moles of each reactant: C4H10: \(10.00 g \cdot \frac{1 \mole}{58.14 g/mol} = 0.172 \mole\) O2: \(10.00 g \cdot \frac{1 \mole}{32.00 g/mol}= 0.3125 \mole\) Find the required ratio of these moles using the stoichiometry of the balanced equation: \( \frac{moles\: C4H10}{2} = \frac{moles\: O2}{13}\) O2 will be limiting if: \( moles\: O2 < \frac{13 \times moles\: C4H10}{2} \) Determine if O2 is limiting: \( 0.3125 < \frac{13 \times 0.172}{2} =0.4465 \) Since 0.3125 < 0.4465, the limiting reactant is oxygen (O2).
03

Calculate the theoretical yield of each product.

Using the limiting reactant (O2) and the stoichiometry of the balanced equation, calculate the theoretical yield of CO2 and H2O: For CO2: \(moles\: CO2 = \frac{8}{13} \times moles\: O2 = \frac{8}{13} \times 0.3125 = 0.1923\,moles\) CONVERT moles to grams: \(0.1923 \,moles \times 44.01 \frac{g}{mol} = 8.466 g\: CO2\) For H2O: \(moles\: H2O = \frac{10}{13} \times moles\: O2 = \frac{10}{13} \times 0.3125 = 0.2404\,moles\) CONVERT moles to grams: \(0.2404 \,moles \times 18.02 \frac{g}{mol} = 4.329 g\: H2O\) The theoretical yield for each product is 8.466 g CO2 and 4.329 g H2O.
04

Determine the grams of excess reactant remaining.

Since we know O2 is the limiting reactant, we can find the amount of C4H10 remaining after the reaction: Determine the moles of C4H10 used: \(\frac{2}{13} \times moles\: O2 = \frac{2}{13} \times 0.3125 = 0.04808 \,moles \) CONVERT moles to grams: \(0.04808 \,moles \times 58.14 \frac{g}{mol} = 2.794 g\) C4H10 remaining: \(10.00 g - 2.794 g = 7.206 g\) There are 7.206 g C4H10 remaining after the reaction.
05

Calculate the additional grams of limiting reactant required.

To determine the additional grams of O2 required for the reaction to be balanced, calculate the moles of O2 needed to react with the remaining C4H10: Initial moles of C4H10: \( \frac{10.00 g}{58.14 \frac{g}{mol}} = 0.172\, moles\) Since oxygen is the limiting reactant: Additional O2 moles: \(\frac{13 \times moles\: C4H10}{2} - moles\: O2 = \frac{13 \times 0.172}{2} - 0.3125 = 0.134\, moles \) Now, convert moles of O2 to grams: Additional O2 required: \(0.134 \,moles \times 32.00 \frac{g}{mol} = 4.288 g\) In order to run the reaction in a balanced fashion, an additional 4.288 g of O2 is required.

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